 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum            Solutions for Session5, Part A

See solutions for Problems: A1 | A2 | A3 | A4 | A5    Problem A1

 a. The number y must be either 5 or 0. Since the sum is not 0, y is 5 and m is 1. b. The digits m and a must be consecutive digits, since (in the tens place) m plus carry equals a. So a is 9 and m is 8. c. Since the sum is two digits, the digit l must be 1 or 2. It cannot be 1, since the sum of four identical numbers is always even. So l is 2, and n must be 3 (since 3 • 4 = 12), and g is 9. d. Since x plus carry equals ba, x must be 9 and ba must be 10. So x is 9, b is 1, and a is 0 (the sum is 999 + 1 = 1,000).   Problem A2 The number a2 is a two-digit number, with digits different from a. The number a3 is a three-digit number, with different digits from both a and a2. Here are the possibilities: a = 5; a2 = 25; a3 = 125 (no, since c = a = 5) a = 6; a2 = 36; a3 = 216 (no, since c = a = 6) a = 7; a2 = 49; a3 = 343 (no, since d = f = 3) a = 8; a2 = 64; a3 = 512 (yes!) a = 9; a2 = 81; a3 = 729 (no, since a = f = 9) The value of bc - a = 64 - 8 = 56.   Problem A3

 a. The solution is 169 • 7 = 1,183. It is easiest to first find the upper-right asterisk, then use the known carry digits to fill in the rest of the product. b. The solution is 47 • 9 = 423. The units-digit asterisk must be 9, since 63 is the only multiple of 7 that ends in 3. Then, with the carry digit 6, only 4 • 9 = 36 + 6 = 42 can give the leading digit of 4. c. The solution is 64 • 6 = 384. The digit being multiplied by must be either 5 or 6 (to give a hundreds digit of 3); if it is 5, the units-digit would have to be 0 or 5. Since it is 4, the digit being multiplied by must be 6. Then the remaining units-digit asterisk must be either 9 or 4; 69 • 6 = 414 is not valid, so the remaining asterisk is 4.   Problem A4 The fact that the sum of a and a equals c, a single digit, means that a can be no more than 4. The fact that the sum of c and b is two different values means that b + c must be larger than 10; it also means that a and b are consecutive numbers. Since the sum of b + c (plus any carry digit) equals b, then c must be either 0 or 9. Knowing that b + c is greater than 10 means that c must equal 9. So the sum is now: Finding a is next. Since a + a (plus any carry digit) equals 9, a must be 4. Then b must be 5, since 9 + b = a, with no carry possibility. The final sum is:    Problem A5 We can solve this by building abcde starting with e. Since e • 3 ends in 1, e must be 7. Then d • 3 + 2 (carry) = 7, so d is 5. Then c • 3 + 1 (carry) = 5, so c is 8. Then b • 3 + 2 (carry) = 8, so b is 2. Then a • 3 + 0 (no carry) = 2, so a is 4. The number abcde is 42,857, and the equation is 142,857 • 3 = 428,571.     